Ethyl acetate is formed by the reaction between ethanol and acetic acid and the equilibrium is represented as:

Exercises · 6.18

Q. Ethyl acetate is formed by the reaction between ethanol and acetic acid and the equilibrium is represented as: CH3COOH (l) + C2H5OH (l) ⇌ CH3COOC2H5 (l) + H2O (l)

(i) Write the concentration ratio (reaction quotient), Qc, for this reaction (note: water is not in excess and is not a solvent in this reaction)

(ii) At 293 K, if one starts with 1.00 mol of acetic acid and 0.18 mol of ethanol, there is 0.171 mol of ethyl acetate in the final equilibrium mixture. Calculate the equilibrium constant.

(iii) Starting with 0.5 mol of ethanol and 1.0 mol of acetic acid and maintaining it at 293 K, 0.214 mol of ethyl acetate is found after sometime. Has equilibrium been reached?

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Concept understanding — Equilibrium Constant Calculation

Equilibrium Constant Calculation: From Intuition to Precision

Imagine you're at a party where people can move between two rooms. Some people prefer the kitchen (more snacks), others prefer the living room (better music). After a while, the number of people in each room stops changing — not because everyone froze, but because the rate of people leaving the kitchen equals the rate of people entering it. The system is in dynamic equilibrium.

Chemical reactions work the same way. A reversible reaction like

$A + B \rightleftharpoons C + D$ doesn't stop when it reaches equilibrium. Instead, the forward reaction (making C and D) and the reverse reaction (making A and B) happen at the same rate. The concentrations of A, B, C, and D become constant — not equal, but constant.

The equilibrium constant $K$ is a number that tells you where this balance lies. It answers the question: At equilibrium, which side of the reaction is favoured?

The Intuitive Idea

Think of a seesaw. If $K$ is very large (say $10^6$ ), the equilibrium sits heavily on the product side — almost all A and B have turned into C and D. If $K$ is very small (say $10^{-6}$ ), the opposite is true: hardly any product forms. If $K$ is around 1, both sides have comparable amounts.

So $K$ is a ratio — a comparison of product concentrations to reactant concentrations at equilibrium, each raised to the power of their stoichiometric coefficients.

The Precise Statement

For a general reversible reaction at a given temperature:

aA + bB \rightleftharpoons cC + dD

the equilibrium constant $K_c$ (for concentrations in mol/L) is:

K_c = \frac{[C]^c [D]^d}{[A]^a [B]^b}

where $[X]$ means the equilibrium concentration of species X in moles per litre.

Important

The exponents come directly from the balanced chemical equation. If the coefficient of A is 2, you square its concentration. This is not optional — it's built into the definition.

What $K$ Actually Depends On

$K$ is constant at a given temperature. Change the temperature, and $K$ changes. But $K$ does not depend on:

Initial concentrations
Presence of a catalyst (catalysts speed up both directions equally)
Pressure or volume changes (for $K_c$ ; $K_p$ for gases has its own rules)

This is a critical exam point: if a problem gives you initial concentrations and asks for $K$ , you must first find equilibrium concentrations — you cannot plug in initial values.

A Worked Example

Problem:

$N_2(g) + 3H_2(g) \rightleftharpoons 2NH_3(g)$

At 500°C, equilibrium concentrations are:

$[N_2] = 0.50\ \text{M}$ , $[H_2] = 1.50\ \text{M}$ , $[NH_3] = 0.20\ \text{M}$

Calculate $K_c$ .

Solution:

Write the expression:

K_c = \frac{[NH_3]^2}{[N_2][H_2]^3}

Substitute:

K_c = \frac{(0.20)^2}{(0.50)(1.50)^3} = \frac{0.04}{0.50 \times 3.375} = \frac{0.04}{1.6875} \approx 0.0237

Note

The units cancel because the numerator and denominator both have units of (mol/L) $^2$ and (mol/L) $^4$ respectively — but by convention, $K_c$ is reported without units. The numerical value is what matters.

Common Pitfalls

Forgetting the exponents: A coefficient of 2 means square the concentration, not double it.
Using initial concentrations: You must use equilibrium concentrations only. …

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For a liquid-phase esterification where water is a product (not a solvent), the equilibrium constant is $K_c = \frac{[\text{ester}][\text{water}]}{[\text{acid}][\text{alcohol}]}$ . Using initial moles and the change table, $K_c \approx 4.0$ . Comparing the reaction quotient $Q_c \approx 3.2$ with $K_c$ shows equilibrium has not been reached — the reaction will proceed forward.

The key insight: this is a homogeneous liquid-phase reaction with no solvent. Water is a product, not a diluent, so it appears in the equilibrium expression. The reaction quotient and equilibrium constant are both written in terms of molar concentrations, but because the volume of the reaction mixture is the same for all species, we can work directly with moles — the volume factor cancels in the ratio.

Let’s go step by step.

1. Writing the reaction quotient $Q_c$

The reaction is:

\text{CH}_3\text{COOH (l)} + \text{C}_2\text{H}_5\text{OH (l)} \rightleftharpoons \text{CH}_3\text{COOC}_2\text{H}_5\text{(l)} + \text{H}_2\text{O (l)}

Since all species are in the same liquid phase and water is not in excess (it’s a product, not the solvent), the concentration ratio is:

Q_c = \frac{[\text{CH}_3\text{COOC}_2\text{H}_5][\text{H}_2\text{O}]}{[\text{CH}_3\text{COOH}][\text{C}_2\text{H}_5\text{OH}]}

Q_c = \frac{[\text{ester}][\text{water}]}{[\text{acid}][\text{alcohol}]}

If water were the solvent (present in huge excess), its concentration would be nearly constant and would be absorbed into $K$ , but here it’s a stoichiometric product — so it stays.

2. Calculating $K_c$ from the first experiment

We start with:

Acetic acid: $1.00$ mol
Ethanol: $0.18$ mol
Ethyl acetate: $0$ mol
Water: $0$ mol

At equilibrium, we are told: ethyl acetate = $0.171$ mol.

Because the stoichiometry is 1:1:1:1, the change in moles is the same for all species. Let $x$ be the moles of ester formed at equilibrium. Then:

Ester formed: $x = 0.171$ mol
Water formed: also $x = 0.171$ mol
Acid consumed: $x = 0.171$ mol → remaining acid = $1.00 - 0.171 = 0.829$ mol
Ethanol consumed: $x = 0.171$ mol → remaining ethanol = $0.18 - 0.171 = 0.009$ mol

Now, the equilibrium constant in terms of moles (since volume $V$ is the same for all):

K_c = \frac{(\text{moles of ester}/V)(\text{moles of water}/V)}{(\text{moles of acid}/V)(\text{moles of alcohol}/V)} = \frac{(\text{moles of ester})(\text{moles of water})}{(\text{moles of acid})(\text{moles of alcohol})}

So:

K_c = \frac{0.171 \times 0.171}{0.829 \times 0.009}

Calculate stepwise:

Numerator: $0.171 \times 0.171 = 0.029241$
Denominator: $0.829 \times 0.009 = 0.007461$
$K_c = \frac{0.029241}{0.007461} \approx 3.92$

Note

Rounding to two significant figures (based on the given data: 0.18, 0.171, etc.), $K_c \approx 3.9$ or $4.0$ . We’ll keep $K_c \approx 4.0$ for comparison.

3. Checking if the second mixture is at equilibrium

New initial amounts: …

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Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

Showing the 12 most recent of 24 on this concept.

CBSE 2026Set sz1 markMCQ
$Q. Select the correct one: The rate at which a substance reacts, depends on its:$
(a) Active mass
(b) Molecular mass
(c) Equivalent mass
(d) Total volume
›Reveal solutionSolution

The Law of Mass Action states that the rate of a chemical reaction is proportional to the product of the active masses (molar concentrations) of the reactants, so the rate depends on active mass.

The Law of Mass Action (proposed by Guldberg and Waage) states that the rate at which a substance reacts is directly proportional to its active mass, where active mass is defined as the molar concentration of the substance (expressed in mol/L, i.e. [reactant]).

For a general reaction aA + bB -> products, the rate is proportional to [A]^a[B]^b — the exponents being the active masses raised to powers related to the stoichiometry (for elementary reactions).

…

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$Q. Give the name of mixture of reactants and products in the equilibrium state.$
›Reveal solutionSolution

At chemical equilibrium, the combined mixture of reactants and products present (in constant, unchanging proportions) is called the equilibrium mixture.

In a reversible reaction, as the reaction proceeds, the forward reaction rate decreases and the reverse reaction rate increases, until both become equal — this is the state of dynamic equilibrium. At this point, the concentrations of all reactants and products stop changing (though the forward and reverse reactions contin …

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CBSE 2026Set ANNUAL1 markMCQ
$Q. Who enunciated the law of mass action ?$
(a) Gulberg and Waage
(b) Bodenstein
(c) Berthelot
(d) Graham
›Reveal solutionSolution

The law of mass action was given by Guldberg and Waage.

Cato Guldberg and Peter Waage (1864) stated the law of mass action: the rate of a chemical reaction is proportional to the product of the active masses (molar concentrations) of the reactants, each raised to the power of its coefficie …

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CBSE 2026Set ANNUAL1 markMCQ
$Q. For which of the following reactions Kp > Kc ?$
(a) N2(g) + 3H2(g) ⇌ 2NH3(g)
(b) H2(g) + I2(g) ⇌ 2HI(g)
(c) PCl3(g) + Cl2(g) ⇌ PCl5(g)
(d) 2SO3(g) ⇌ 2SO2(g) + O2(g)
›Reveal solutionSolution
Kp > Kc requires Δn(gas) > 0; only 2SO3 ⇌ 2SO2 + O2 satisfies this.

The relation is Kp = Kc(RT)^Δn, where Δn = (moles of gaseous products) − (moles of gaseous reactants). Kp > Kc needs Δn > 0.

(A) N2 + 3H2 ⇌ 2NH3: Δn = 2 − 4 = −2 → Kp < Kc.

(B) H2 + I2 ⇌ 2HI: Δn = 2 − 2 = 0 → Kp = Kc. …
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CBSE 2026Set ANN1 markMCQ
$Q. Equilibrium constant (Kc) of a reaction depends on :$
(a) temperature
(b) pressure
(c) catalyst
(d) initial concentration
›Reveal solutionSolution
Kc depends only on temperature; it is unaffected by pressure, catalyst or initial concentration. Answer: (a).

The equilibrium constant is a fixed number for a given reaction at a given temperature.

Changing pressure or initial concentration shifts the position of equilibrium but does NOT change Kc. …
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CBSE 2025Set ANNUAL1 markMCQ
$Q. In which of the following is Kp less than Kc?$
(a) PCl5 ⇌ PCl3 + Cl2
(b) H2 + Cl2 ⇌ 2HCl
(c) 2SO2 + O2 ⇌ 2SO3
(d) All of these
›Reveal solutionSolution

Kp < Kc only for the reaction where the number of gas moles decreases (Δn negative): 2SO2 + O2 ⇌ 2SO3.

The relation is Kp = Kc(RT)^Δn, where Δn = (moles of gaseous products) - (moles of gaseous reactants). Since RT > 1 (in atm-litre units, for reasonable temperatures), Kp > Kc when Δn is positive, Kp = Kc when Δn = 0, and Kp < Kc when Δn is negative.

Checking each reaction: …

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CBSE 2025Set ANNUAL1 markMCQ
$Q. If 64 gram of Hydrogen iodide are dissolved in 2 litre of water then its active mass will be$
(a) 0.5
(b) 0.25
(c) 1.0
(d) 2.5
›Reveal solutionSolution

64 g of HI in 2 L of water gives an active mass (concentration) of 0.25 mol/L.

Molar mass of HI = 1 + 127 = 128 g/mol.

Moles of HI = 64 g / 128 g mol^-1 = 0.5 mol.

…

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CBSE 2025Set ANNUAL1 markMCQ
$Q. For the following chemical reaction PCl3(g) + Cl2(g) ⇌ PCl5(g). The value of Kc at 250°C is 26 then the value of Kp at this temperature will be$
(a) 0.61
(b) 0.57
(c) 0.83
(d) 0.46
›Reveal solutionSolution

For PCl3(g) + Cl2(g) ⇌ PCl5(g) at 250°C, Kp ≈ 0.61.

Δn = (moles of gaseous product) - (moles of gaseous reactants) = 1 - 2 = -1.

T = 250°C + 273 = 523 K. R = 0.0821 L atm K^-1 mol^-1.

…

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CBSE 2025Set ANNUAL1 markMCQ
$Q. For a reversible reaction A + B (equilibrium arrows) C + D, the equilibrium constant Kc is equal to$
(a) [A][B] / [C][D]
(b) [C][A] / [B][D]
(c) [C][D] / [B][A]
(d) [A][D] / [B][C]
›Reveal solutionSolution

Kc = (concentrations of products, each raised to its coefficient) / (concentrations of reactants, each raised to its coefficient) — for A + B <=> C + D, that's [C][D]/[A][B].

For a general reversible reaction at equilibrium:

aA + bB <=> cC + dD

The law of mass action gives the equilibrium constant expression as:

Kc = [C]^c [D]^d / ([A]^a [B]^b)

Here the reaction is A + B <=> C + D (all coefficients = 1), so:

Kc = [C][D] / [A][B]

…

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CBSE 2025Set ANNUAL1 markMCQ
$Q. If the value of Kc for the equilibrium 2NOCl(g) <=> 2 NO(g) + Cl2(g) is 9.0 x 10^-4 mol L^-1 then numerical value of Kc for the equilibrium NOCl(g) <=> NO(g) + 1/2 Cl2(g) will be.................$
(a) 4.5 x 10^-4
(b) 3.0 x 10^-4
(c) 4.5 x 10^-2
(d) 3.0 x 10^-2
›Reveal solutionSolution

Halving all coefficients of a balanced equation converts K to sqrt(K). Since the given Kc = 9.0x10^-4 is for the doubled reaction, the halved reaction's Kc is sqrt(9.0x10^-4) = 3.0x10^-2.

For the reaction 2NOCl(g) <=> 2NO(g) + Cl2(g):

Kc = [NO]^2[Cl2] / [NOCl]^2 = 9.0x10^-4

The second reaction, NOCl(g) <=> NO(g) + 1/2 Cl2(g), is exactly half of the first reaction (all coefficients divided by 2).

…

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$Q. Case study: A chemist while studying a number of equilibria found that there was a relationship between Kp and Kc. He tested this relation upon various equilibria at different temperatures and the relation was found to be true. He further noticed that for certain equilibria Kp and Kc were equal, however it was not always true. Based on above study answer the following question:$
(b) Under what condition Kp and Kc are equal?
›Reveal solutionSolution

From Kp = Kc(RT)^delta-n, the two constants become equal exactly when the exponent delta-n is zero, i.e., when there is no change in the total number of moles of gas between reactants and products.

From the derived relation, Kp = Kc(RT)^delta-n, where delta-n = (moles of gaseous products) - (moles of gaseous reactants).

If delta-n = 0 (i.e., the number of moles of gaseous species is the same on both sides of the balanced equation), then (RT)^delta-n = (RT)^0 = 1, and the relation simplifies to:

Kp = Kc

…

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CBSE 2025Set ANNUAL1 mark
$Q. Case study: A chemist while studying a number of equilibria found that there was a relationship between Kp and Kc. He tested this relation upon various equilibria at different temperatures and the relation was found to be true. He further noticed that for certain equilibria Kp and Kc were equal, however it was not always true. Based on above study answer the following question:$
(c) Write the relation between Kp and Kc for the equilibrium 2H2(g) + O2(g) <=> 2H2O(g)
›Reveal solutionSolution

With delta-n = 2 - (2+1) = -1 for this reaction, the general relation Kp = Kc(RT)^delta-n becomes Kp = Kc(RT)^-1, i.e., Kp = Kc/(RT).

For the equilibrium: 2H2(g) + O2(g) <=> 2H2O(g)

Count gaseous moles:

Reactants: 2 mol H2 + 1 mol O2 = 3 mol of gas

Products: 2 mol H2O = 2 mol of gas

delta-n = (moles of gaseous products) - (moles of gaseous reactants) = 2 - 3 = -1

Using the general relation Kp = Kc(RT)^delta-n:

Kp = Kc(RT)^(-1) = Kc / (RT)

…

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Q.Ethyl acetate is formed by the reaction between ethanol and acetic acid and the equilibrium is represented as: CH3COOH (l) + C2H5OH (l) ⇌ CH3COOC2H5 (l) + H2O (l)

1. Writing the reaction quotient QcQ_cQc​

2. Calculating KcK_cKc​ from the first experiment

3. Checking if the second mixture is at equilibrium

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1. Writing the reaction quotient $Q_c$

2. Calculating $K_c$ from the first experiment

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